Light from a point source in air falls on a spherical glass surface ($n = 1.5$ and radius of curvature $R = 20 \ cm$). The distance of the light source from the glass surface is $100 \ cm$. Find the image distance. (in $cm$)

$A$ spherical convex surface of power $5 \text{ dioptre}$ separates object and image space of refractive indices $1.0$ and $\frac{4}{3}$ respectively. The radius of curvature of the surface is (in $\text{ cm}$)

$A$ spherical surface of radius of curvature $R$ separates air (refractive index $n_1 = 1.0$) from glass (refractive index $n_2 = 1.5$). The center of curvature is inside the glass. $A$ point object placed at $P$ forms a real image at $Q$. The line $PQ$ intersects the surface at $O$ and $PO = OQ$. The distance $PO$ is:

In the figure shown here,a point object $O$ is placed in air. $A$ spherical boundary separates various media with a radius of curvature $2.0 \, m$. $AB$ is the principal axis. The refractive index above $AB$ is $1.6$ and below $AB$ is $2.0$. The separation between the images formed due to refraction at the spherical surface is

$A$ spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. $A$ point object $O$ is placed in air on the optic axis of the surface,so that its real image is formed at $I$ inside the glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ is equal to: (in $R$)

$A$ hemispherical glass lens with a refractive index of $1.5$ is placed in a liquid with a refractive index of $1.3$ (see the figure). The radius of the hemispherical lens is $10 \,cm$. $A$ parallel beam of light traveling in the liquid is refracted by the glass lens. The absolute value of the position of the image from the center of the glass lens will be (in $\,cm$)